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inna [77]
3 years ago
5

Ed wants to buy a car for $17,500. His Credit union approves him for a car loan with a 6.8% APR as long as you puts 20% down. He

is looking into the following options.
1) How much money would Ed have to put down to qualify for the loan?


2) What would the loan amount be?

Option A: Pay $400 per month.

3) What would be the length of his loan? Round to the nearest tenth of a year.

4) What is the total amount of interest Ed would pay over the life of this loan?

Option B: Pay $350 per month.

5) What would be the length of his loan? Round to the nearest tenth of a year.

6) What is the total amount of interest Ed would pay over the life of this loan?
Mathematics
1 answer:
kkurt [141]3 years ago
5 0

Answer:

Thank you for the freeeee points.

Step-by-step explanation:

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PLEASE EEEEE !!!!!!!!!!!! I NEED IT
tino4ka555 [31]

Answer:

1) ∠A=84°

2) ∠C=20°

Step-by-step explanation:

1)

First, find ∠C:

<em>(I'm assuming the exterior angle of 126° makes a straight line with ∠C)</em>

The angles on a straight line always add up to 180. Therefore:

∠C+126=180

∠C=180-126

∠C=54

Then find ∠B:

We also know that all the angles in a triangle add up to 180. Therefore:

∠A+∠B+∠C=180

∠A+∠B+54=180

∠A+∠B=126

<em>(we know ∠A=2(∠B))</em>

2(∠B)+∠B=126

3(∠B)=126

∠B=42

Now, find ∠A:

∠A=2(∠B)

∠A=2(42)

∠A=84°

2)

First, find ∠B:

<em>(Again, I'm assuming the exterior angle of 100° makes a straight line with ∠B)</em>

The angles on a straight line always add up to 180. Therefore:

∠B+100=180

∠B=180-100

∠B=80

Then find ∠A:

We also know that all the angles in a triangle add up to 180. Therefore:

∠A+∠B+∠C=180

∠A+80+∠C=180

∠A+∠C=100

<em>(we know ∠A=4(∠C))</em>

4(∠C)+∠C=100

5(∠C)=100

∠C=20°

6 0
3 years ago
Please help. The correct and best answer gets brainiest! 29 points!!
attashe74 [19]

Answer:(15, 100)

Step-by-step explanation:

first find the coordinates of the point above (3, 20)

you just look at the x axis and where the point is in this case x would be 6 and then you find y which would be 40

so the coordinate would be (6, 40)

now subtract (6, 40) by (3, 20)

6-3= 3

40- 20= 20

you get the first coordnites which is what you add (12,80) to

12 +3= 15

80+ 20= 100

you get the coordinates (15, 100)

4 0
3 years ago
NEED HELP ASAP PLZZZZZZZZZZZZZZZ
alisha [4.7K]

Answer: I would say B would be the answer Because....

Step-by-step explanation:

if about 20 to 23 cars pass through every 26 minutes, and haw many cars pass by in 156 minutes...

26 x 6 = 156

6 x 20 = 120  

6 x 21 = 126

6 x 22 = 132

6x 23 = 138

all thoes numbers are around B.

4 0
3 years ago
Find the value of x.
V125BC [204]

Answer: B. 14

Step-by-step explanation:

because 140 divided by 10 equals 14

3 0
3 years ago
Read 2 more answers
Math<br><br><br><br> pls help!!<br><br><br><br><br><br> answers?
statuscvo [17]

Answer: Choice B) Infinitely many solutions

  • one solution: x = 8, y = -7/2, z = 0
  • another solution: x = -12, y = 13/2, z = 10

=======================================================

Explanation:

Here's the starting original augmented matrix.

\left[\begin{array}{ccc|c}  1 & 0 & 2 & 8\\5 & 1 & 9 & 73/2\\-4 & 0 & -8 & -32\\\end{array}\right]

We'll multiply everything in row 3 (abbreviated R3) by the value -1/4 or -0.25, which will make that -4 in the first column turn into a 1.

We use this notation to indicate what's going on: (-1/4)*R3 \to R3

That notation says "multiply everything in R3 by -1/4, then replace the old R3 with the new corresponding values".

So we have this next step:

\left[\begin{array}{ccc|c}  1 & 0 & 2 & 8\\5 & 1 & 9 & 73/2\\1 & 0 & 2 & 8\\\end{array}\right]\begin{array}{l}  \ \\\ \\(-1/4)*R3 \to R3\\\end{array}

Notice that the new R3 is perfectly identical to R1.

So we can subtract rows R1 and R3, and replace R3 with the result of nothing but 0's

\left[\begin{array}{ccc|c}  1 & 0 & 2 & 8\\5 & 1 & 9 & 73/2\\0 & 0 & 0 & 0\\\end{array}\right]\begin{array}{l}  \ \\\ \\R3-R1 \to R3\\\end{array}

Whenever you get an entire row of 0's, it <u>always</u> means there are infinitely many solutions.

-------------------

Now let's handle the second row. That 5 needs to turn into a 0. We can multiply R1 by 5, and subtract that from R2.

So we need to compute 5*R1-R2 and have that replace R2.

\left[\begin{array}{ccc|c}  1 & 0 & 2 & 8\\0 & 1 & -1 & -7/2\\0 & 0 & 0 & 0\\\end{array}\right]\begin{array}{l}  \ \\5*R1-R2 \to R2\ \\\ \\\end{array}

Notice that in the third column of R2, we have 9-5*2 = 9-10 = -1. So we have -1 replace the 9. In the fourth column of R2, we have 73/2 - 5*8 = -7/2. So the -7/2 replaces the 73/2.

--------------------

At this point, the augmented matrix is in RREF form. RREF stands for Reduced Row Echelon Form. It seems a bit odd that the "F" of "RREF" stands for "form" even though we say "form" right after "RREF", but I digress.

Because the matrix is in RREF form, this means R1 and R2 lead to these equations:

R1 : 1x+0y+2z = 8\\ R2: 0z+1y-1z = -7/2

which simplify to

R1: x+2z = 8\\R2: y-z = -7/2

Let's get the z terms to each side like so:

x+2z = 8\\x = -2z+8\\\text{ and }\\y-z = -7/2\\y = z-7/2\\

Therefore, all of the solutions are of the form (x,y,z) = (-2z+8, z-7/2, z) where z is any real number.

If z is allowed to be any real number, then we can simply pick any number we want to replace it. We consider z to be the "free variable", in that it's free to be whatever it wants. The values of x and y will depend on what we pick for z.

So the concept of "infinitely many solutions" doesn't exactly mean we can pick just <em>any</em> triple for x,y,z (admittedly it would be nice to randomly pick any 3 numbers off the top of my head and be done right away). Instead, we can pick anything we want for z, and whatever we picked, will directly determine x and y. The x and y are locked into place so to speak.

Let's say we picked z = 0.

That would lead to...

x = -2z+8\\x = -2(0)+8\\x = 8\\\text{ and }\\y = z-7/2\\y = 0-7/2\\y = -7/2\\

So z = 0 would lead to x = 8 and y = -7/2

Rearranging the items in alphabetical order gets us:

x = 8, y = -7/2, z = 0

We have one solution of (x,y,z) = (8, -7/2, 0)

Now let's say we picked z = 10

x = -2z+8\\x = -2(10)+8\\x = -12\\\text{ and }\\y = z-7/2\\y = 10-7/2\\y = 13/2\\

So we have x = -12, y = -13/2, z = 10

Another solution is (x,y,z) = (-12, 13/2, 10)

There's nothing special about z = 0 or z = 10. You can pick any two real numbers you want for z. Just be sure to recalculate the x and y values of course.

To verify each solution, you'll need to plug them back into the original equations formed by the original augmented matrix. After simplifying, you should get the same thing on both sides.

8 0
3 years ago
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